Question

Two satellites have their masses in the ratio 3:1. The radii of their circular orbits are in the ratio 1:4. What is the total mechanical energy of A and B?​

Answer 1

PLEASE SEE THE ATTACHMENT

Answer 2

Answer :

›»› The ratio of total mechanical energy of A and B are 12 : 1.

Given :

• Two satellites have their masses in the ratio 3:1. The radii of their circular orbits are in the ratio 1:4.

To Find :

• What is the ratio of total mechanical energy of A and B?

Solution :

Total mechanical energy :

$\displaystyle{\tt{\longmapsto \frac{ - GMm}{2r}}}$

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Here we are provided that,

$\tt{\longmapsto \dfrac{m_1}{m_2} = \dfrac{3}{1} }$

$\tt{\longmapsto \dfrac{r_1}{r_2} = \dfrac{1}{4} }$

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So, now their total energy (E) is :

$\displaystyle{\tt{\longmapsto E_A\frac{ - GMm}{2r_{1}}}}$

$\displaystyle{\tt{\longmapsto E_B \frac{ - GMm}{2r_{2}}}}$

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Now,

Their ratio will be,

$\displaystyle{\tt{: \implies \dfrac{ \dfrac{ \not - GMm_{1}}{2r_{1}} }{\dfrac{ \not - GMm_{2}}{2r_{2}}}}}$

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$\displaystyle{\tt{: \implies \dfrac{ \dfrac{ \not{G} \not{M}m_{1}}{2r_{1}} }{\dfrac{ \not{G} \not{M}m_{2}}{2r_{2}}}}}$

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$\displaystyle{\tt{: \implies \dfrac{ \dfrac{m_{1}}{ \not{2}r_{1}} }{\dfrac{m_{2}}{ \not{2}r_{2}}}}}$

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$\displaystyle{\tt{: \implies \dfrac{ \dfrac{m_{1}}{r_{1}} }{\dfrac{m_{2}}{r_{2}}}}}$

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$\displaystyle{\tt{: \implies \frac{m_{1}r_{2}}{m_{2}r_{1}} }}$

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$\displaystyle{\tt{: \implies \frac{3 \times 4}{1 \times 1} }}$

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$\displaystyle{\tt{: \implies \frac{12}{1 \times 1} }}$

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$\displaystyle{\bf{: \implies \frac{12}{1} }}$

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Hence, the ratio of total mechanical energy of A and B are 12 : 1.